Solve the system of linear equations using matrices. x+y+z=3 2x+3y+2z=7 3x-4y+z=4

Step 1
Given equation
x+y+z=3
2x+3y+2z=7
3x-4y+z=4
in matrix form
$A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 3& -4& 1\end{array}\right]$
$B=\left[\begin{array}{c}3\\ 7\\ 4\end{array}\right]$
$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
we know that $A\cdot X=B$
So, $X=A^{-1}\cdot B$
Step 2
Now,
$X=A^{-1}\cdot B$
We have to find $A^{-1}$
Adjoin the identity matrix onto the right of the original matrix, so that you have A on the left side and the identity matrix on the right side. $\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 3& -4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
now we will apply the elementary row method $R_2-2R_1\to R_2$ (multiply 1 row by 2 and subtract it from 2 row), $R_3-3R_1\to R_3$ (multiply 1 row by 3 and subtract it from 3 row)
$$\begin{gathered} \left[\begin{array}{ccc}1& 1& 1\\ 2-2\cdot 1& 3-2\cdot 1& 2-2\cdot 1\\ 3-3\cdot 1& -4-3\cdot 1& 1-3\cdot 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0-2\cdot 1& 1-2\cdot 0& 0-2\cdot 0\\ 0-3\cdot 1& 0-3\cdot 0& 1-3\cdot 0\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 0& -7& -2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ -3& 0& 1\end{array}\right] \end{gathered}$$
$R_1-1R_2\to R_1$ (multiply 2 row by 1 and subtract it from 1 row), $R_3+7R_2\to R_3$ (multiply 2 row by 7 and add it to 3 row)
$\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& -2\end{array}\right]=\left[\begin{array}{ccc}3& -1& 0\\ -2& 1& 0\\ -17& 7& 1\end{array}\right]$
$\frac{R_3}{-2}\to R_3$ (divide the 3 row by -2)
$\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& -1& 0\\ -2& 1& 0\\ 8.5& -3.5& -0.5\end{array}\right]$
$R_1-1R_3\to R_1$ (multiply 3 row by 1 and subtract it from 1 row)
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-5.5& 2.5& 0.5\\ -2& 1& 0\\ 8.5& -3.5& -0.5\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}-5.5& 2.5& 0.5\\ -2& 1& 0\\ 8.5& -3.5& -0.5\end{array}\right]$
Step 3
Now,
$X=A^{-1}\cdot B$

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